__Exercise 13.4 Page No: 225__

__Exercise 13.4 Page No: 225__

**1. Find the surface area of a sphere of radius:**

**(i) 10.5cm (ii) 5.6cm (iii) 14cm**

**(Assume π=22/7)**

**Solution**:

Formula: Surface area of sphere (SA) = 4πr^{2}

(i) Radius of sphere, r = 10.5 cm

SA = 4×(22/7)×10.5^{2 }= 1386

Surface area of sphere is 1386 cm^{2}

(ii) Radius of sphere, r = 5.6cm

Using formula, SA = 4×(22/ 7)×5.6^{2 }= 394.24

Surface area of sphere is 394.24 cm^{2}

(iii) Radius of sphere, r = 14cm

SA = 4πr^{2}

= 4×(22/7)×(14)^{2}

= 2464

Surface area of sphere is 2464 cm^{2}

**2. Find the surface area of a sphere of diameter:**

**(i) 14cm (ii) 21cm (iii) 3.5cm**

**(Assume π = 22/7)**

**Solution:**

(i) Radius of sphere, r = diameter/2 = 14/2 cm = 7 cm

Formula for Surface area of sphere = 4πr^{2}

= 4×(22/7)×7^{2} = 616

Surface area of a sphere is 616 cm^{2}

(ii) Radius (r) of sphere = 21/2 = 10.5 cm

Surface area of sphere = 4πr^{2}

= 4×(22/7)×10.5^{2 }= 1386

Surface area of a sphere is 1386 cm^{2}

Therefore, the surface area of a sphere having diameter 21cm is 1386 cm^{2}

(iii) Radius(r) of sphere = 3.5/2 = 1.75 cm

Surface area of sphere = 4πr^{2}

= 4×(22/7)×1.75^{2} = 38.5

Surface area of a sphere is 38.5 cm^{2}

**3. Find the total surface area of a hemisphere of radius 10 cm. [Use π=3.14]**

**Solution:**

Radius of hemisphere, r = 10cm

Formula: Total surface area of hemisphere = 3πr^{2}

= 3×3.14×10^{2} = 942

The total surface area of given hemisphere is 942 cm^{2}.

**4. The radius of a spherical balloon increases from 7cm to 14cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.**

**Solution:**

Let r_{1 }and r_{2} be the radii of spherical balloon and spherical balloon when air is pumped into it respectively. So

r_{1 }= 7cm

r_{2 }= 14 cm

Now, Required ratio = (initial surface area)/(Surface area after pumping air into balloon)

= 4r_{1}^{2}/4r_{2}^{2}

= (r_{1}/r_{2})^{2}

= (7/14)^{2 }= (1/2)^{2} = ¼

Therefore, the ratio between the surface areas is 1:4.

**5. A hemispherical bowl made of brass has inner diameter 10.5cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm ^{2}. (Assume π = 22/7)**

**Solution:**

Inner radius of hemispherical bowl, say r = diameter/2 = (10.5)/2 cm = 5.25 cm

Formula for Surface area of hemispherical bowl = 2πr^{2}

= 2×(22/7)×(5.25)^{2} = 173.25

Surface area of hemispherical bowl is 173.25 cm^{2}

Cost of tin-plating 100 cm^{2} area = Rs 16

Cost of tin-plating 1 cm^{2} area = Rs 16 /100

Cost of tin-plating 173.25 cm^{2 }area = Rs. (16×173.25)/100 = Rs 27.72

Therefore, the cost of tin-plating the inner side of the hemispherical bowl at the rate of Rs 16 per 100 cm^{2} is Rs **27.72.**

**6. Find the radius of a sphere whose surface area is 154 cm ^{2}. (Assume π = 22/7)**

**Solution:**

Let the radius of the sphere be r.

Surface area of sphere = 154 (given)

Now,

4πr^{2 }= 154

r^{2 }= (154×7)/(4×22) = (49/4)

r = (7/2) = 3.5

The radius of the sphere is 3.5 cm.

**7. The diameter of the moon is approximately one fourth of the diameter of the earth.**

**Find the ratio of their surface areas.**

**Solution:**

If diameter of earth is said d, then the diameter of moon will be d/4 (as per given statement)

Radius of earth = d/2

Radius of moon = ½×d/4 = d/8

Surface area of moon = 4π(d/8)^{2}

Surface area of earth = 4π(d/2)^{2}

The ratio between their surface areas is 1:16.

**8. A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5cm. Find the outer curved surface of the bowl. (Assume π =22/7)**

**Solution:**

Given:

Inner radius of hemispherical bowl = 5cm

Thickness of the bowl = 0.25 cm

Outer radius of hemispherical bowl = (5+0.25) cm = 5.25 cm

Formula for outer CSA of hemispherical bowl = 2πr^{2}, where r is radius of hemisphere

= 2×(22/7)×(5.25)^{2} = 173.25

Therefore, the outer curved surface area of the bowl is 173.25 cm^{2}.

**9. A right circular cylinder just encloses a sphere of radius r (see fig. 13.22). Find**

**(i) surface area of the sphere,**

**(ii) curved surface area of the cylinder,**

**(iii) ratio of the areas obtained in(i) and (ii).**

**Solution:**

(i) Surface area of sphere = 4πr^{2}, where r is the radius of sphere

(ii) Height of cylinder, h = r+r =2r

Radius of cylinder = r

CSA of cylinder formula = 2πrh = 2πr(2r) (using value of h)

= 4πr^{2}

(iii) Ratio between areas = (Surface area of sphere)/CSA of Cylinder)

= 4r^{2}/4r^{2 }= 1/1

Ratio of the areas obtained in (i) and (ii) is 1:1.